·\'11 Verizon 5:29 PM 62% session.masteringchemistry.com , A sample of sodium-24

Question

·'11 Verizon 5:29 PM 62% session.masteringchemistry.com , A sample of sodium-24 with an activity of 10 mCi is used to study the rate of blood flow in the circulatory system. Sodium-24 has a half-life of 15 h You may want to reference (Pages 165-168) Section 5.4 while completing this problem Part A What is the radioactivity of the sample of sodium-24 after one half-life? Express the activity to two significant figures and include the appropriate units. alue Units Part B What is the radioactivity of the sample of sodium-24 after 30 h? Express the activity to two significant figures and include the appropriate units ale Units Part C What is the radioactivity of the sample of sodium-24 after three half-lives? Express your answer to two significant figures and include the appropriate units. Value Units Part D What is the radioactivity of the sample of sodium-24 after 2.5 days? Express the activity to two significant figures and include the appropriate units

Explanation / Answer

part A

first order,

k = 0.693/t1/2

   = 0.693/15

   = 0.0462 s-1

k = (1/t)ln(a0/a)

a0 = Initial activity of Na-24 = 10 mCi

a = activity of Na-24 after one half-life = x mCi

t = one half life = 15 h

0.0462 = (1/15)ln(10/x)

x = 5 mCi

part B

first order,

k = 0.693/t1/2

   = 0.693/15

   = 0.0462 s-1

k = (1/t)ln(a0/a)

a0 = Initial activity of Na-24 = 10 mCi

a = activity of Na-24 after one half-life = x mCi

t = 30 h

0.0462 = (1/30)ln(10/x)

x = 2.5 mCi

part C

k = (1/t)ln(a0/a)

a0 = Initial activity of Na-24 = 10 mCi

a = activity of Na-24 after one half-life = x mCi

t = 3*15 = 45 h

0.0462 = (1/45)ln(10/x)

x = 1.25 mCi

part D

k = (1/t)ln(a0/a)

a0 = Initial activity of Na-24 = 10 mCi

a = activity of Na-24 after one half-life = x mCi

t = 2.5*24 = 60 h

0.0462 = (1/60ln(10/x)

x = 0.625 mCi

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